# Question #bcf2d

Dec 30, 2016

$\textsf{16.2 \textcolor{w h i t e}{x} \text{m/s}}$

#### Explanation:

See fig. (a)

First we find the vertical component of the motion using:

$\textsf{{v}^{2} = {u}^{2} + 2 a s}$

This becomes:

$\textsf{{v}_{y}^{2} = 0 + 2 \times g \times h}$

$\textsf{{v}_{y}^{2} = 0 + 2 \times 9.81 \times 6 = 117.72}$

$\textsf{{v}_{y} = \sqrt{117.72} = 10.85 \textcolor{w h i t e}{x} \text{m/s}}$

See fig (b).

The horizontal component of velocity is constant so we can find the resultant velocity $\textsf{{V}_{r e s}}$ using Pythagoras:

$\textsf{{12}^{2} + {10.85}^{2} = {V}_{r e s}^{2}}$

$\therefore$$\textsf{{V}_{r e s}^{2} = 261.72}$

$\textsf{{V}_{r e s} = \sqrt{261.72} = 16.2 \textcolor{w h i t e}{x} \text{m/s}}$

If you were asked to find the angle to the vertical then:

$\textsf{\tan \theta = \frac{12}{10.85} = 1.105}$

From which $\textsf{\theta = {48}^{\circ}}$